(x-1)^2 * (x+2)^3 >= 0
helpppp please!
Well, the left side cannot be negative, (or zero)can it?
The first term can never be negative, but the second can if x<-2
So as long as x is >-2, the statement is true.
oops, the statement includes zero as being allowed, so x>=-2
To solve the inequality (x-1)^2 * (x+2)^3 ≥ 0, we can follow these steps:
Step 1: Solve for critical points
First, we need to determine the critical points by setting each factor equal to zero.
Setting (x-1)^2 = 0:
(x-1)(x-1) = 0
x - 1 = 0
x = 1
Setting (x+2)^3 = 0:
(x+2)(x+2)(x+2) = 0
x + 2 = 0
x = -2
So the critical points are x = 1 and x = -2.
Step 2: Create a number line
Next, we'll create a number line and plot the critical points on it. This will help us determine the intervals where the inequality is true.
-∞ -2 1 +∞
---------------|--------------------
Step 3: Choose a test point in each interval
Select a value from each interval and plug it into the inequality to see if it is true. We can use the number line to guide our choices.
Let's choose the test points x = -3, 0, and 2.
For x < -2 (interval: -∞ to -2), let's plug in x = -3:
(-3 - 1)^2 * (-3 + 2)^3 ≥ 0
(-4)^2 * (-1)^3 ≥ 0
16 * (-1) ≥ 0
-16 ≥ 0
This inequality is false, so the solution is not in this interval.
For -2 < x < 1 (interval: -2 to 1), let's plug in x = 0:
(0 - 1)^2 * (0 + 2)^3 ≥ 0
(-1)^2 * (2)^3 ≥ 0
1 * 8 ≥ 0
8 ≥ 0
This inequality is true, so the solution is in this interval.
For x > 1 (interval: 1 to +∞), let's plug in x = 2:
(2 - 1)^2 * (2 + 2)^3 ≥ 0
(1)^2 * (4)^3 ≥ 0
1 * 64 ≥ 0
64 ≥ 0
This inequality is true, so the solution is in this interval.
Step 4: Determine the solution
Based on the test points, we find that the inequality holds true for the interval -2 < x < 1 and for x > 1. Therefore, the solution to the inequality (x-1)^2 * (x+2)^3 ≥ 0 is:
x ∈ (-∞, -2] ∪ [1, +∞)